hdu 2586 How far away ？（LCA 求两点距离）

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3252    Accepted Submission(s): 1222

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

 

Input

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

 

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

 

Sample Input

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

题意：给出一个无向带权图，给出m个询问a b，对于每个询问，求a到b的距离。

思路：参考了http://www.cnblogs.com/ylfdrib/archive/2010/11/03/1867901.html。lca是离线算法，在求dfs的过程中把根节点到各节点的距离保存起来，最后对于每两点（a,b）之间的距离，d=dis[a]+dis[b]-2*dis[lca(a,b)]。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <vector>#include <cmath>#include <map>#include <cstdlib>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll long longusing namespace std;const int maxn=40005;struct node{    int v,w,next;}edge[maxn*2];struct node1{    int u,v,next,id;}query[maxn*2];int G[maxn],Q[maxn],dis[maxn],fa[maxn],lca[500];bool vis[maxn];int n,m,s,e,num,numq,ans;void init(){    memset(G,-1,sizeof(G));    memset(Q,-1,sizeof(Q));    num=numq=0;}void add(int u,int v,int w){    edge[num].w=w;    edge[num].v=v;    edge[num].next=G[u];    G[u]=num++;}void addq(int u,int v,int id){    query[numq].u=u;    query[numq].v=v;    query[numq].id=id;    query[numq].next=Q[u];    Q[u]=numq++;}int Find_set(int x){    return x==fa[x]?x:fa[x]=Find_set(fa[x]);}void input(){    int a,b,c;    scanf("%d%d",&n,&m);    for(int i=1;i<n;i++)    {        scanf("%d%d%d",&a,&b,&c);        add(a,b,c);        add(b,a,c);    }    for(int i=1;i<=m;i++)    {        scanf("%d%d",&a,&b);        addq(a,b,i);        addq(b,a,i);    }}void dfs(int u){    vis[u]=true;    fa[u]=u;    for(int i=Q[u];i!=-1;i=query[i].next)    {        int v=query[i].v;        if(vis[v]) lca[query[i].id]=Find_set(v);    }    for(int i=G[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(!vis[v])        {            dis[v]=dis[u]+edge[i].w;            dfs(v);            fa[v]=u;        }    }}void solve(){    memset(vis,false,sizeof(vis));    dis[1]=0;    dfs(1);    for(int i=0;i<numq;i+=2)    {        int u=query[i].u;        int v=query[i].v;        int id=query[i].id;        printf("%d\n",dis[u]+dis[v]-2*dis[lca[id]]);    }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        init();        input();        solve();    }    return 0;}